Math 3181
Name:
Dr. Franz Rothe
February 6, 2014
All3181\3181_spr14h2.tex
Homework has to be turned in this handout.
The homework can be done in groups up to three
due February 11/12
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Solution of Homework
Definition 1 (Quadrangle and quadrilateral). Four points, no three of which lie
on a line, are said to be a quadrangle. These four points are necessarily distinct.
Four lines, no three of which intersect at a point, are said to be a quadrilateral.
Again these four lines are necessarily distinct.
Definition 2 (Projective plane). A projective plane is a class of points, and a class
of lines satisfying the axioms:
P.1 Every two points lie on exactly one line.
P.2 Every two lines intersect at exactly one point.
P.3 There exist four points of which no three lie on a line.
Remark. The lines are not necessarily sets of points.
10 Problem 2.1. Which parallel property holds for a projective plane?
Convince yourself that on every line of a projective plane lie at least three points.
Secondly, convince yourself that in every point intersect at least three lines, and that
a quadrilateral exists.
Answer. By axiom (P2), there holds the elliptic parallel property. By axiom (P.3), there
exist at least four points A, B, C, D of which no three lie on a line. Let any point P
be given. In the case that point P is different from all four points A, B, C, D, we draw
the lines P A, P B, P C and P D. At least three of them are different since no three
points among A, B, C, D lie on a line. In the case that point P is one of the four points
A, B, C, D, we get three different lines through the given point P among the six lines
connecting A, B, C, D. In both cases we have obtained three different lines through the
arbitrary point P .
The proof confirming that three points lie on every line is done quite similarly,
interchanging the roles of points and lines.
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For Hilbert’s detailed investigations about the Theorem of Desargues, it is enough
to use the following formulation ”in parallel setting”. This is not any real restriction.
By means of the projective completion, one can get back to the general Theorem of
Desargues, spelled out in the script.
Proposition 1 (Theorem of Desargues in parallel setting)..
(i) If two triangles in a plane have three pairs of parallel corresponding sides, then the
three lines through their corresponding vertices either meet in one point, or all
three are parallel.
(ii) If the three lines through the corresponding vertices of two triangles in a plane
either meet in one point, or all three are parallel, and moreover, two pairs of
corresponding sides are parallel, then the third pair of corresponding sides are
parallel, too.
These statements are true for a plane which is part of a three dimensional incidence
geometry, satisfying the axioms of incidence (I.1) through (I.8) and the strict parallel
postulate (IV*).
10 Problem 2.2. Provide a drawing for statement (i) of Proposition 1 above. Use
blue shades for the lines assumed parallel, and red to mark the coincidence asserted.
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10 Problem 2.3. Provide a drawing for statement (ii) of Proposition 1 above.
Use now blue shades for the coincidence and the two pairs of parallel assumed, and red
to mark the parallelism asserted.
Figure 1: The Theorem of Desargues in parallel setting.
Answer.
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Definition 3 (The projective plane over a field F). The "points" of the projective
plane PF2 are the sets of equivalent triples (xλ, yλ, zλ), where x, y, z ∈ F and λ runs
over the nonzero elements of F. The "lines" of the projective plane are equations
ax + by + cz = 0
with coefficients a, b, c ∈ F, not all three equal to zero.
A "point lies on a line" if and only if the coordinate triple (x, y, z) satisfies the
equation of the line.
The triples (x, y, z) ∈ F3 \ (0, 0, 0) are the homogeneous coordinates for the points of
the projective plane. Similarly, the triples (a, b, c) ∈ F3 \ (0, 0, 0) are the homogeneous
coordinates for the lines of the projective plane.
Again, it is easy to check, but important to confirm:
Main Theorem 1. The projective plane over any field is a projective plane.
10 Problem 2.4 (Three points on a line). In a projective coordinate plane
three points with the homogeneous coordinates (x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) are
given. Check that the three points lie on a line if and only if the determinant
x1 x2 x3 y1 y2 y3 = 0
z1 z2 z3 You may use vectors or three-dimensional analytic geometry, to keep the argument as
simple as possible.
Answer. The three points of the projective plane with the homogeneous coordinates
(x1 , y1 , z1 ), (x2 , y2 , z2 ) and (x3 , y3 , z3 ) lie on a line if and only if the three lines through
the origin
{(λx1 , λy1 , λz1 ) : λ ∈ F}
{(λx2 , λy2 , λz2 ) : λ ∈ F}
{(λx3 , λy3 , λz3 ) : λ ∈ F}
lie in a plane in F3 . It is known that this happens if and only if the determinant of their
coordinates from above is zero.
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10 Problem 2.5. Give exact definitions for segment and for ray. Include the
obviously necessary details. Use only the primary notions.
Answer.
Definition 4 (Segment). Let A and B be two distinct points. The segment AB is the
set consisting of the points A and B and all points lying between A and B. The points
A and B are called the endpoints of the segment, the points between A and B are called
the interior points, and the remaining points on the line AB are called the exterior
points of the segment.
−→
Definition 5 (Ray). Given two distinct points A and B, the ray AB is the set consisting
of the points A and B, the points inside the segment AB, and all points P on the line
AB such that the given point B lies between A and P . The point A is called the vertex
of the ray.
10 Problem 2.6. Give an exact definition for a triangle. Use the primary notions. Include the obviously necessary details.
Answer.
Definition 6 (Triangle). We define a triangle to be the union of the three segments
AB, BC and CA. The three points A, B and C are assumed not to lie on a line. These
three points are the vertices, and the segments BC, AC, and AB are the sides of the
triangle.
Remark. For the congruence, similarity of two triangles, and for the Theorem of Desargues, one has to consider the vertices and sides of these triangles in a definite order.
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Definition 7. Let A, B and C be three points on the given line with B lying between A
−→
−−→
and C. The two rays BA and BC. are called the opposite rays if their common vertex
B lies between A and C. We say that two points in the same ray lie on the same side
of the vertex. Two points in opposite rays lie on different sides of the vertex.
Proposition 2 (”Line separation Theorem”). Given is a line and a vertex lying on
it. Each point of the line except the vertex is contained in exactly one of the opposite
rays originating from the vertex.
Proposition 3 (Hilbert’s Proposition 4, also called ”Three-point Theorem”).
Among any three points A, B and C lying on a line, there exists exactly only one lying
between the two other points.
Proposition 4 (Hilbert’s Proposition 5, also called ”Four-point Theorem”).
Any four points on a line can be notated in a way that all four order relations that keep
the alphabetic order hold.
10 Problem 2.7. Why does the line separation theorem and the three point theorem, together imply the four-point theorem. Complete the following reasoning.
Reasoning. Given are four different points on a line l. By the !! Three-point Theorem
, one of them, which we name B lies between two others. Some of the four given points
lie on different sides of B. Denote by A the point lying by itself on one side of B. The
points on the other side are !! C and D . We name them such that B ∗ C ∗ D holds.
We have thus obtained the three order relations
!! A ∗ B ∗ C , A ∗ B ∗ D , B ∗ C ∗ D
We still have to confirm the fourth order relation, concerning A, C, D. All three points
B, C, D lie on one ray emanating from A. And all three points !! A, B, C lie on one
ray emanating from D. Hence only one order relation, namely A ∗ C ∗ D is possible.
Thus in the list A ∗ B ∗ C ∗ D, all four alphabetic order relations hold.
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Remark. The Line separation Theorem does not follow neither simply from the axioms
(II.1)(II.2) and (II.3), nor from the Three-point Theorem.
The following counterexample has been constructed by Hartshorne. The "points"
are the integers modulo 5. Hence there exist just five points. There is one "line"
through all five points. The "order relation" is defined by requiring
(2.1)
a ∗ b ∗ c if and only if
2b ≡ a + c (mod 5)
Question. Convince yourself that
axioms (II.1) and (II.2) are valid for this new ordering;
of any three distinct points 0, a, b exactly one lies between the two others. There are
six cases to be checked:
(0, a, b) = (0, 1, 4), (0, 2, 3), (0, 1, 2), (0, 2, 4), (0, 1, 3), (0, 3, 4)
In each case, find the middle point and write down the valid order relation.
Show that of any three distinct points, exactly one lies between the other two.
Answer. Axiom (II.1) is valid for this new ordering. Indeed, a ∗ b ∗ c implies c ∗ b ∗ a
since a + c = c + a.
Axiom (II.2) is valid. Given two points a and c, let b :≡ 2c−a (mod 5). Since a+b ≡ 2c
(mod 5), we get a point b (beyond the segment) such that point c lies between a
and b.
Of any three distinct points 0, a, b exactly one lies between the two others. Indeed, for
the six possible cases, the order turns out to be:
1∗0∗4, 2∗0∗3, 0∗1∗2, 0∗2∗4, 0∗3∗1, 0∗4∗3
Given any three distinct points x, y, z, let a := y − x and b := z − x. We order the
three points 0, a, b and finally get for the three points x, y, z the corresponding
order relation.
10 Problem 2.8. To see that the line separation Theorem does not follow neither
simply from the axioms (II.1)(II.2) and (II.3), nor from the Three-point Theorem, we
proceed as follows. Convince yourself that for the order defined above
a ∗ x ∗ c if and only if
a ∗ c ∗ y if and only if
x ≡ 3(a + c) (mod 5)
y ≡ 2c − a (mod 5)
For better reading I denote the five points now by P0 , P1 , P2 , P3 , P4 .
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• Of which three points consists the segment P0 P2 .
−−→
• Of which four points consists the ray P0 P2 .
• Of which points consists the segment P0 P3 .
−−→
• Of which points consists the ray P0 P3 .
−−→
−−→
• Why are P0 P2 and P0 P3 two opposite rays.
−−→
−−→
• Which points are common to the opposite rays P0 P2 and P0 P3 .
• Why is the line separation Theorem not valid in this interpretation.
• The segment P0 P2 consists of the points P0 , P1 , P2 .
−−→
• The ray P0 P2 consists of the points P0 , P1 , P2 , P4 .
Answer.
• The segment P0 P3 consists of the points P0 , P4 , P3 .
−−→
• The ray P0 P3 consists of the points P0 , P4 , P3 , P1 .
−−→
−−→
• Since P2 ∗ P0 ∗ P3 , the rays P0 P2 and P0 P3 are opposite rays.
−−→
−−→
• The opposite rays P0 P2 and P0 P3 have the points P0 ,P1 and P4 in common.
• Since the two opposite rays have more points than their common vertex in common, the line separation theorem is violated.
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10 Problem 2.9. We consider points and lines in one plane. What does it mean
to say that the points A and A0 lie on the same side of line a. What does it mean to
say points A and B lie on different sides of line a.
Answer. We say that points A and A0 lie on the same side of line a iff the segment AA0
does not intersect the line a. We say that points A and B lie on different sides of line a
iff the segment AB does intersect the line a.
Proposition 5 (Hilbert’s Proposition 8, the ”Plane separation Theorem”).
Each line a, lying in a plane α, separates the points of this plane which do not lie on
the line, into two nonempty regions R and S called half planes which have the following
properties:
(i) every point lying in the plane α lies either on the line a, or in the half plane R, or
in the halfplane S, but not in any two of these three sets;
(ii) for any two points A in the halfplane R and B in the halfplane S, the segment AB
intersects line a;
(iii) for any two points A and A0 in the halfplane R, the segment AA0 does not intersect
the line a;
(iv) for any two points B and B 0 in the half plane S, the segment BB 0 does not intersect
the line a.
The following proof of the plane separation theorem uses only the axioms of order
and Bernay’s Lemma.
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Figure 2: Generic cases in the proof of the plane separation theorem.
Figure 3: Special cases in the proof of the plane separation theorem.
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below.
Problem 2.10. Complete the proof of the plane separation theorem given
A simple proof of the plane separation theorem. By axiom !! (I.3a) , there exist two
points P 6= Q on the line a. By axiom (I.3b), there exists a third point R in the plane
α, not lying on the line a. The three points P, Q and R span the plane α.
We can now separate the points of the plane α into those lying on the line a and,
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additionally, the following two sets:
R : = {A ∈ α : A ∈
/ a, the segment RA does not intersect the line a}
S : = {B ∈ α : B ∈
/ a, the segment RB intersects the line a}
By axiom (II.2), there exist a point S such that !! R ∗ P ∗ S . In other words, the
−→
point S lies on the extension of the ray RP beyond point P . Similarly, we get point T
such that R ∗ Q ∗ T . Since the points P, Q and R do not lie on a line, neither the three
points R, S and T lie on a line.
Since R ∈ R, and S, T ∈ S, both sets are !! nonempty . From the definition, it is
clear that each point of the plane α lies in exactly one of the three sets R, a or S. Thus
!! item (i) has been checked.
In the special case that A = R or A0 = R, the statements (ii) and (iii) follow from
the definition. Moreover, we use repeatedly the following statement:
Lemma 1. Assume that three points A ∈ R and B ∈ S and R as above lie on a line r,
but point S does not lie on this line. Then line a intersects segment AS, but does not
intersect segment BS.
Reason for the Lemma. In the triangle 4ASR, the line a intersects the side SR by the
construction of point S, but does not intersect side AR. By Pasch’s axiom the line a
intersects !! the third side AS , as to be shown.
In the triangle 4BSR, the line a intersects both sides SR and BR. By !! Bernay’s Lemma
, the line a cannot intersect all three sides of a triangle. Hence the line a does not intersect the third side !! BS , as to be shown.
We now assume that A 6= R and A0 6= R, and check statements (ii),(iii) and (iv).
(ii) Given are a point A in the region R and a point B in the region S. We distinguish
two cases:
• The three points R, A and B do not lie on a line. We apply Pasch’s axiom
to the triangle !! 4RAB and line a. The assumptions A ∈ R and B ∈ S
mean that line a does !! not intersect the side AR , but does intersect the
side BR of the triangle. By Pasch’s axiom the line a intersects a second side
which can only be !! side AB , as to be shown.
• We consider now the special case that points R, A and B lie on a line. Either
the three points S, A and B or the three points T, A and B do not lie on
a line. Since the roles of S and T can be switched, I need only to consider
the first possibility. We use Pasch’s axiom for triangle 4ABS. Since line a
intersects the side AS but not the side SB, the line a intersects the side AB,
as to be shown.
(iii) Given are two points A and A0 in the region R. We distinguish two cases:
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• The three points R, A and A0 do not lie on a line. We apply Pasch’s axiom
to the triangle !! 4RAA0 and line a. The assumptions A ∈ R and A0 ∈ R
mean that line a does not intersect neither the side AR nor the side A0 R. By
Pasch’s axiom the line a !! cannot intersect the third side AA0 , as to be
shown.
• We consider now the special case that points R, A and A0 lie on a line. Either
the point S or point T does not lie on this line. Since the roles of S and T
can be switched, I need only to consider the first possibility.
We use Bernay’s Lemma for triangle 4AA0 S. Since line a intersects both
sides AS and A0 S, the line a does not intersect the side AA0 , as to be shown.
(iv) Given are two points B and B 0 in the region S. We distinguish two cases:
• The three points R, B and B 0 do not lie on a line. We apply Pasch’s axiom to the triangle !! 4RBB 0 and line a. The assumptions B ∈ S and
B 0 ∈ S mean that line a does not intersect both sides BR and B 0 R. By
!! Bernay’s Lemma , the line a cannot intersect the third !! side BB 0 , as
to be shown.
• We consider now the special case that points R, B and B 0 lie on a line. Either
the point S or point T does not lie on this line. Since the roles of S and T
can be switched, I need only to consider the first possibility.
We use Pasch’s axiom for triangle !! 4BB 0 S . Since line a does not intersect
neither side BS nor !! side B 0 S , the line a does not intersect the side BB 0 ,
as to be shown.
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Figure 4: A ray interior of an angle intersects a segment from side to side.
Proposition 6 (The Crossbar Theorem). A segment with endpoints on the two sides
of an angle and a ray emanating from its vertex into the interior of the angle intersect.
10 Problem 2.11. Give a proof for the Crossbar Theorem. Complete the figure
on page 13 according to your proof.
Figure 5: The Crossbar Theorem is proved using Pasch’s axiom.
−→
Answer. Let ray r = AC lie in the interior of the given angle ∠BAD, and let B and D
−→
be arbitrary points on the two sides of this angle. We have to show that the ray AC
intersects the segment BD.
−→
Let F be any point on the ray opposite to AB. We apply Pasch’s axiom to triangle
4F BD and line l = AC. The line intersects the side F B of the triangle in point A,
and does not pass through neither one of the vertices F, B, D. We check that side F D
does not intersect line AC.
−→
Indeed, the points inside segment F D and the points inside ray AC lie on different
sides of line AD. But the points inside segment F D and inside the opposite ray lie on
different sides of line AB. (Points A, D, F are exceptions, but neither can segment F D
and line AC intersect in any of these points.)
Hence—by Pasch’s axiom—the third side BD intersects line AC, say at point Q.
Segment BD, and hence the intersection point Q are in the interior of ∠BAD. Since
−→
only the ray AC, but not its opposite ray, lies in the interior of ∠BAD, the intersection
−→
point Q lies on the ray AC. Here is a drawing, to show how Pasch’s axiom is applied.
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